Answer
$\frac{\lambda_{electron}}{\lambda_{proton}}=1.8362\times10^{3}$
Work Step by Step
Given that electron and proton are having the same speed $v$
mass of the electron is $m_{e}= 9.109\times^{-31}kg$
speed of the electron is $v$
so momentum of electron $p_{e}=m_{e}v$
suppose de_Broglie wavelength of the electron is $\lambda_{electron}$
so from de Broglie hypothesis $\lambda=\frac{h}{p}$
$\lambda_{electron}=\frac{h}{p_{e}}=\frac{h}{m_{e}v}$........................equation(1)
mass of the proton is $m_{p}= 1.6726\times^{-27}kg$
speed of the proton is $v$
so momentum of proton $p_{p}=m_{p}v$
suppose de_Broglie wavelength of the electron is $\lambda_{electron}$
so from de Broglie hypothesis $\lambda=\frac{h}{p}$
$\lambda_{proton}=\frac{h}{p_{p}}=\frac{h}{m_{p}v}$........................equation(2)
dividing equation(1) by equation(2)
$\frac{\lambda_{electron}}{\lambda_{proton}}$=$\frac{\frac{h}{m_{e}v}}{\frac{h}{m_{p}v}}$
$\frac{\lambda_{electron}}{\lambda_{proton}}=\frac{m_{p}}{m_{e}}$
putting
$m_{e}= 9.109\times^{-31}kg$
$m_{p}= 1.6726\times^{-27}kg$
$\frac{\lambda_{electron}}{\lambda_{proton}}=\frac{1.6726\times^{-27}kg}{9.109\times^{-31}kg}$
$\frac{\lambda_{electron}}{\lambda_{proton}}=0.18362\times10^{4}$
$\frac{\lambda_{electron}}{\lambda_{proton}}=1.8362\times10^{3}$
